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2 years ago

8

A softball fits snuggly inside a cubical box that is 4 inches on each side. What is the volume of the softball that fits snuggly

inside the box? Use 3.14 for π. Round to the nearest tenth, if necessary.

1 answer:

Natali5045456 [20]2 years ago

4 0

Answer:

33.5 cubic inches.

Step-by-step explanation:

If the softball fits perfectly inside the cubical box, the softball is a sphere with diameter of 4 inches (radius os 2 inches).

The volume of a sphere is given by:

$V= \frac{4}{3}\pi r^3$

For r = 2 inches:

$V= \frac{4}{3}*3.14* 2^3\\V=33.5\ in^3$

The volume of the softball is 33.5 cubic inches.

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A toy manufacturer wants to know how many new toys children buy each year. A sample of 305 children was taken to study their pur

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Answer:

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

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Now, find the margin of error M as such

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In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

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The lower end of the interval is the sample mean subtracted by M. So it is 7.6 - 0.1 = 7.5

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The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

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Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

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Answer:

a

$\= x = 18.5$ , $\sigma = 5.15$

b

$15.505 < \mu < 21.495$

c

$14.93 < \mu < 22.069$

Step-by-step explanation:

From the question we are are told that

The sample data is 21, 14, 13, 24, 17, 22, 25, 12

The sample size is n = 8

Generally the ample mean is evaluated as

$\= x = \frac{\sum x }{n}$

$\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}$

$\= x = 18.5$

Generally the standard deviation is mathematically evaluated as

$\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}$

$\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}$

$\sigma = 5.15$

considering part b

Given that the confidence level is 90% then the significance level is evaluated as

$\alpha = 100-90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.645$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.645 * \frac{5.15 }{\sqrt{8} }$

=> $E = 2.995$

The 90% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 2.995 < \mu < 18.5 + 2.995$

$15.505 < \mu < 21.495$

considering part c

Given that the confidence level is 95% then the significance level is evaluated as

$\alpha = 100-95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.96 * \frac{5.15 }{\sqrt{8} }$

=> $E = 3.569$

The 95% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 3.569 < \mu < 18.5 + 3.569$

$14.93 < \mu < 22.069$

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